Excercises Electric Machinery Fundamentals

Chapter 6 Problem 6-6



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

For the motor in Problem 6-5

What is the slip at the pullout torque?
What is the pullout torque of this motor?



In [2]:

    
R1 =  0.10     # [Ohm]
R2 =  0.07     # [Ohm]
Xm = 10.0      # [Ohm]
X1 =  0.21     # [Ohm]
X2 =  0.21     # [Ohm]
Pmech  = 500   # [W]
Pmisc  =   0   # [W]
Pcore  = 400   # [W]
Vphi   = 120   # [V]
w_sync = 188.5 # [rad/s]

SOLUTION

The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply, and then using that with the rotor circuit model.

$$Z_{TH} = \frac{jX_M(R_1+jX_1)}{R_1 + j(X_1+X_M)}$$



In [3]:

    
Zth = (Xm*1j * (R1 + X1*1j)) / (R1 + (X1+Xm)*1j)
Zth_angle = arctan(Zth.imag/Zth.real)
print('Zth = ({:.4f})Ω = {:.3f} Ω ∠{:.1f}°'.format(Zth, abs(Zth), Zth_angle/pi*180))









    



Zth = (0.0959+0.2066j)Ω = 0.228 Ω ∠65.1°

$$\vec{V_{TH}} = \frac{jX_M}{R_1 + j(X_1+X_M)} \vec{V_ \phi }$$



In [4]:

    
Vth = Xm*1j / (R1 + (X1+Xm)*1j) * Vphi
Vth_angle = arctan(Vth.imag/Vth.real)
print('Zth = {:.1f} V ∠{:.1f}°'.format(abs(Vth), Vth_angle/pi*180))









    



Zth = 117.5 V ∠0.6°

The slip at pullout torque is: $$S_\text{max} = \frac{R_2}{\sqrt{R_{TH}^2+(X_{TH}+X_2)^2}}$$



In [5]:

    
Rth = Zth.real
Xth = Zth.imag
s_max = R2 / sqrt(Rth**2 + (Xth+X2)**2)
print('''
s_max = {:.3f}
============='''.format(s_max))









    



s_max = 0.164
=============

The pullout torque of the motor is:

$$\tau_\text{max} = \frac{3V_{TH}^2}{2\omega_\text{sync}[R_{TH}+\sqrt{R_{TH}^2+(X_{TH}+X_2)^2}]}$$



In [6]:

    
tau_max = (3* abs(Vth)**2) / (2 * w_sync*(Rth + sqrt(Rth**2 + (Xth+X2)**2)))
print('''
tau_max = {:.0f} Nm
================'''.format(tau_max))









    



tau_max = 210 Nm
================